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I am working on the following problem:

Show that if $A$ is free on a set $S$ in Ab, with map $\mathrm{\Phi}:S \rightarrow A$, then $A$ is also a coproduct of $|S|$ copies of $\mathbb{Z}$, where each $\mathrm{\theta}_x :\mathbb{Z} \rightarrow A$ is given by $\mathrm{\theta}_x(n)=n\cdot \mathrm{\Phi}(x)$.

I can't use the fact that $A$ is free on $S$ and the definition of a coproduct to prove that $A$ is isomorphic with the coproduct of $|S|$ copies of $\mathrm{Z}$. How do you construct the diagram? I only have to work with the definitions of free object and coproduct.

Definitions:

$X$ is a set. $F$ with $\phi :X \to F$ is free on $X$ in the category $\mathcal{C}$ iff $\forall A \in \mathcal{O}(\mathcal{C})$ and $\psi :X\to A$, $\exists! f \in Mor(F,A)$ such that $f\circ \phi = \psi$.

Suppose $\{A_i : i\in \mathcal{I}\}$ is an indexed family in $\mathcal{O}(\mathcal{C})$. A coproduct of the $A_i$ is an object $A$ together with morphisms $\iota_i \in Mor(A_i,A)$ for all $i\in \mathcal{I}$ satisfying the property: if $B\in \mathcal{O}(\mathcal{C})$, and $\phi_i \in Mor(A_i,B)$ for all $i\in \mathcal{I}$ then $\exists !\tau:A \to B$ such that all the following diagrams are commutative: $$ \begin{array}[c]{ccc} A_i&\xrightarrow{\iota_i}&A\\ &\llap{\phi_i}\searrow& \big\downarrow\rlap{\tau}\\ &&B \end{array}$$

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    $\begingroup$ It's pretty strange that you must prove something without hypotheses. Every proof coming to my mind would use at the end the universal property of $\coprod$; anyways, you can argue this way: $A\cong F(S)$, where $F\colon {\bf Set}\to {\bf Ab}$ is the free abelian group functor, left adjoint ot the forgetful $U\colon {\bf Ab}\to {\bf Set}$; now since $$ {\bf Ab}(\coprod_{s\in S}\mathbb{Z},B)\cong \prod_{s\in S}B $$ naturally in $B$, since RHS is precisely the set of functions $S\to B$, and by uniqueness of adjoints, you get $F(S)\cong \mathbb{Z}^{(S)}$ $\endgroup$ – Fosco Loregian Jun 3 '16 at 7:03
  • $\begingroup$ Thanks for the answer. I am not very familiar with category theory because I am studying only the basics in my homological algebra class. I have a solution and I would appreciate it if you would check it. $\endgroup$ – user53970 Jun 13 '16 at 17:13
  • $\begingroup$ I changed the problem a little to include the definitions I can use. $\endgroup$ – user53970 Jun 13 '16 at 18:12
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$\begin{array}[c]{ccc} A&\stackrel{\Phi}{\longleftarrow}&S\\ \Big\uparrow\rlap{\phi_x}&\searrow &\Big\downarrow\rlap{\xi}\\ \mathbb{Z}_x&\stackrel{\iota_x}{\longrightarrow}&\bigoplus\mathbb{Z}_x \end{array}$

Lets define $\phi_x \in Mor(\mathbb{Z}_x , A)$, so that $\phi_x(n)=n\cdot\Phi(x)$. Since $\bigoplus\mathbb{Z}_x$ is a coproduct, by definition we have $\Rightarrow \exists !f\in Mor(\bigoplus\mathbb{Z}_x,A)$ such that $f\iota_x=\phi_x$.

Now define $\xi(x)=\iota_x(1)$. Since $A$ is free on $S$, by definition we have $\Rightarrow \exists !g\in Mor(A,\bigoplus\mathbb{Z}_x)$ such that $g\Phi = \xi$.

Lets prove that $A\cong \bigoplus\mathbb{Z}_x$. For this we need that $fg=gf=id_A$:

For all $a\in A$, $\exists x\in S$ such that $\Phi(x)=a$ because $\Phi$ is one-to-one. Hence: $$fg(a)=fg\Phi(x)=f\xi(x)=f\iota_x(1)=\phi_x(1)=\Phi(x)=a$$ We concluded that $fg=id_A$. Now, for all $z\in \bigoplus\mathbb{Z}_x$, $\exists x\in S$ and $\exists n\in\mathbb{Z}_x$ such that $z=\iota_x(n)$. So: $$gf(z)=gf\iota_x(n)=g\phi_x(n)=n\cdot g\Phi(x)=n\cdot\xi(x)=n\cdot \iota_x(1)=\iota_x(n)=z$$ So $gf=id_A$ and with this we are done.

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